3-3 when other suits break evenly

[gwnn]

I had nothing better to do today so I was multiplying and dividing C(n,m). I was wondering, how much does the probability for a 3-3 break change when we know more and more about the hand?

 

Let's say, we know nothing about the hand except that each opponent has 13 cards. In that case, everybody knows the 35.5%-48.4%-14.5%-1.5% proportions for 33-42-51-60. What if we know that each opponent has only N vacant spaces?

 

N	"3-3"	"4-2"	"5-1"	"6-0"
13	0.355	0.484	0.145	0.015
12	0.360	0.485	0.141	0.014
11	0.365	0.487	0.136	0.012
10	0.372	0.488	0.130	0.011
9	0.380	0.489	0.122	0.009
8	0.392	0.490	0.112	0.007
7	0.408	0.490	0.098	0.005
6	0.433	0.487	0.078	0.002
5	0.476	0.476	0.048	0.000
4	0.571	0.429	0.000	0.000
3	1.000	0.000	0.000	0.000   

So even if each opponent has 6 vacant spaces, 4-2 will be more likely than 3-3. Also, up until about 5 cards known (8 vacant spaces), the probabilities crawl up very slowly.

 

What about if we have two different suits, and we need one of them to split 3-3? A simple calculation would give the result of 1-(1-0.355)^2=58.43% - how well is this reproduced by the exact numbers?

 

First, let's just see how the probabilities change (2-4 etc refer to the short hand having the short other suit as well):

split	"3-3"	"2-4"	"4-2"	42/24	"1-5"	"5-1"	51/15	"0-6"	"6-0"	60/06
(prior)	0.355	0.242	0.242	0.484	0.072	0.072	0.145	0.008	0.008	0.015
3-3	0.372	0.244	0.244	0.488	0.065	0.065	0.130	0.005	0.005	0.011
4-2	0.358	0.179	0.307	0.485	0.036	0.107	0.143	0.002	0.012	0.014
5-1	0.318	0.119	0.358	0.477	0.017	0.163	0.181	0.001	0.024	0.025
6-0	0.258	0.070	0.387	0.458	0.007	0.232	0.239	0.000	0.044	0.044

The probability of 3-3 is only a bit better than before when a suit breaks 3-3, while it is about the same as a priori when a suit breaks 4-2 (it's bad news that the other suit created a small imbalance but it's good news that it wasn't 5-1 or 6-0 so as to create a bigger imbalance). Also, interestingly, the 35.5% still only decreases to 31.8% when a suit breaks 5-1. The small chance that a suit is 6-0 increases about threefold when another suit is 6-0 but less than twofold when another suit is 5-1.

 

OK so what about the main question I asked?

 

The exact probability of at least one suit breaking 3-3 is 57.86%. So the "naive" 58.43% is not far off at all. If we need both suits to break 3-3, the "naive" way of calculating gives us 12.62%, while the exact result is 13.20%. Again, naive works.

 

I know all of these numbers are on a website somewhere but I haven't seen it discussed a lot. I guess that is because the naive way works just fine.

 

anyway sorry for taking up your time (and -what is infinitely worse- my own).

[manudude03]

I think people just do the naive way of calculating simply because the other way is a lot harder, and the naive way is a decent approximation. The way to get the a priori chance of a 3-3 break is (deliberately non simplified for clarity):

 

13/26 * 12/25 * 11/24 * 13/23 * 12/22 * 11/21 * 6!/(3!*3!) which gives 0.3552795031

 

If we were to check for a 3-3 break given 10 vacant spaces each would become:

 

10/20 * 9/19 * 8/18 * 10/17 * 9/16 * 8/15 * 6!/(3!*3!)

 

The above gives an answer of 0.37151702786, but who has the time to actually do the above calculation in their head at the table?

[gwnn]

Of course nobody has the time to do it, I was just wondering by how much it actually changes. I noticed that most people even on the forums use the "naive" way and I was a bit puzzled by it until today when I actually took the half an hour needed to confirm that the simple multiplication of probabilities gives very decent approximations unless something drastic happened (5-1, 6-0, or a lot of known cards).

 

Also it's funny how resilient the 4-2 percentages are. 4-2 vs 2-4 is obviously changing but except for a single entry (4-4 vacant spaces), it is always somewhere between 46%-49% and almost always even closer. 3-3 loses some power and the uneven ones gain some when breaks get more uneven, but 4-2 soldiers on.

[jogs]

Of course nobody has the time to do it, I was just wondering by how much it actually changes. I noticed that most people even on the forums use the "naive" way and I was a bit puzzled by it until today when I actually took the half an hour needed to confirm that the simple multiplication of probabilities gives very decent approximations unless something drastic happened (5-1, 6-0, or a lot of known cards).

 

 

It is a hypergeometric distribution. This is a math and stat function under f_x in EXCEL.

[gwnn]

Well yea. I got them in Excel too. I know the function but didn't have a feeling for how fast it changes. Now I do: not very fast.

[jogs]

When one suit breaks 4-2, the other suit is much more likely to break 2-4 than 4-2.

[rhm]

When one suit breaks 4-2, the other suit is much more likely to break 2-4 than 4-2.

very brilliant.

We did not expect that

 

Rainer Herrmann

[gwnn]

very brilliant.

We did not expect that

 

Rainer Herrmann

A very good criticism of the whole thread, I think. :lol:

(not sarcastic)